Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?

Solution

Given,
Energy of the electron beam = 20 MeV

Energy is given by,

\frac{1}{2} {mv}^{2} = 20 Mev = 20 \times 1.6 \times 10^{- 13} J

Therefore,

velocity, v = \sqrt{\frac{2 \times 20 \times 1.6 \times 10^{- 13}}{9.1 \times 10^{- 31}}} m / s = 2.65 \times 10^{9} m / s

The velocity of the electron beam is greater than velocity of light. Thus, we can say that the electron is clearly moving at relativisitic speed.

So, the non-relativistic formula

r = \frac{m_{0} v}{eB}

is not valid.

We should use relativistic formula:

Thus,
Radius of the path is given by,

r = \frac{mv}{eB}

That is ,

r = \frac{m_{0} v}{eB \sqrt{1 - \frac{v^{2}}{c^{2}}}} .

where, m_o is the relativistic mass.

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Wave Optics

Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?

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